Suppose men\'s heights are normally distributed with mean 69.2 inches and standa
ID: 3178801 • Letter: S
Question
Suppose men's heights are normally distributed with mean 69.2 inches and standard deviation 2.9
inches.
Answer the following using a calculator and the table:
a) What range of heights are 68.27% of men between?
b) What percentage of men are within 2 standard deviations of the mean?
c) What percentage of men are between 5 foot, 6 inches and 6 feet tall?
d) What is the probability that a randomly selected man is more than 6 feet tall?
3. Now re-do the previous problem using R. Write down any differences. (Give the code in R)
Explanation / Answer
Given that the mean (u) of a normal probability distribution is 69.2 and the
standard deviation (sd) is 2.9
a.
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [69.2 ± 2.9]
= [ 69.2 - 2.9 , 69.2 + 2.9]
= [ 66.3 , 72.1 ]
b.
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [69.2 ± 2 * 2.9]
= [ 69.2 - 2 * 2.9 , 69.2 + 2* 2.9]
= [ 63.4 , 75 ]
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 66) = (66-69.2)/2.9
= -3.2/2.9 = -1.1034
= P ( Z <-1.1034) From Standard Normal Table
= 0.13492
P(X < 72) = (72-69.2)/2.9
= 2.8/2.9 = 0.9655
= P ( Z <0.9655) From Standard Normal Table
= 0.83286
P(66 < X < 72) = 0.83286-0.13492 = 0.6979
d.
P(X > 72) = (72-69.2)/2.9
= 2.8/2.9 = 0.9655
= P ( Z >0.966) From Standard Normal Table
= 0.1671